Problem Solving Strategies

Koolman

2:45 AM

In this velocity of end B is 0

so how the pulses meet at $AB/\sqrt 2$

Anonymous

@Koolman Velocity of pulse produced at end B is not 0.

Anonymous

Find out tension in the string as a function of distance $x$

Anonymous

$v=\sqrt{\frac{T}{\mu}}$

Anonymous

5:27 AM

Wait a minute

Anonymous

Uploading the question

Anonymous

Is the question visible?

Anonymous

@Yashas

Yashas

loading

Anonymous

5:30 AM

Calculate $\Delta H$ for one mole of a real gas undergoing a change in state from $10^5$ Pa, 300 K to $2*10^5$ Pa, 600 K, $a\rightarrow 0$

Yashas

oh nvm I had to click on the image

loaded

the preview takes time to load

Anonymous

b=50 ml/mole

Anonymous

Cp=20

Yashas

H = U + PV

dH = dU + d(PV)

Anonymous

No no I know that

Anonymous

5:31 AM

The problem is in the solution

Anonymous

Pressure and temperature

Anonymous

Are both variable

Anonymous

How do I integrate ?

Anonymous

Lemme show you the solution

Yashas

dU is always equal to nCvdT

dH is always equal to nCpdT

Anonymous

5:31 AM

@Yashas Nooooooooooooooooooooooooooooooooooooo

Yashas

dH = q when P is constant

Anonymous

That is for ideal gas only

Anonymous

This is a real gas

Yashas

that works for real gas too

Anonymous

No

Yashas

5:32 AM

:|

Anonymous

It doesn't

Yashas

it works for solids too :|\

Anonymous

I'm showing you the formula for real gases

Yashas

and liquids

Anonymous

It doesn't

Yashas

5:32 AM

why shouldn't it work for real gas?

Anonymous

You're wrong

Yashas

o0

Anonymous

For real gases

Anonymous

And here is the correct formula

Anonymous

Lemmme type

Anonymous

5:33 AM

wait a min

Anonymous

@Yashas See the formula in the picture

Anonymous

There is an additional term

Anonymous

In dH

Anonymous

For real gases and solids

Yashas

5:34 AM

the first term is dU

the second is d(PV)

Anonymous

There is an additional term (so this statement was wrong)

Anonymous

3 mins ago, by Yashas

dH is always equal to nCpdT

Yashas

the pic has to load

Anonymous

dH neq to nCpdT

Anonymous

For real gases

Anonymous

5:36 AM

Even dU is not equal to only nCvdT

Yashas

dU = dKE + dPE

Anonymous

@Yashas Yeah, you didn't say that earlier

Yashas

equipartition of energy considers both kinds of energy

so you can fit it into one Cv

Anonymous

You just said $dU=nC_vdT$

Yashas

yea

Cv has fix for the PE

Anonymous

5:39 AM

Cp can't account for change in pressure..

Yashas

I am confused now

Anonymous

That dP term has to be accounted for

Yashas

I have been using that eqn for solids and liquids

for solids and liquids, the volume change is ignored

so d(PV) = VdP

what's P for a solid and liquid?

Anonymous

@Yashas I'm talking about real gases. And secondly dP and dV may not be close to 0 for all solids/liquids

Anonymous

Although we assume it

Anonymous

5:41 AM

@Yashas External pressure

Yashas

U is the sum of all internal energies

let me start from there

it includes KE and PE

Temperature measures the kinetic energy or the internal energy?

Anonymous

Internal energy

Anonymous

U is a function of both volume and temperature

Yashas

I think I need to do a revision of thermo

@blue I think the second term is from the fact that Cp changes over pressure/tempearture

I am not sure

Anonymous

@Yashas yes

Yashas

5:55 AM

if Cp wasn't changing, then it wud be nCpdT always?

Anonymous

@Yashas If pressure weren't changing..

Anonymous

then yes

Anonymous

That dP=0

Yashas

if pressure was changing and Cp was independent of P, then also it wud be equalk to nCpdT

I am so confused now

need to do a quick revision

I have confusion about internal energy itself

Sid

7:55 AM

2

Q: How to evaluate this limit? Riemann Integral

Find the coordinates of the maxima of $f(x)$ First you take a natural logarithm and simplify to get an expression and then notice a Riemann SUM. Simplify this and you will get an expression independent of "n". The expression will depend only on x. Please help me out to find this and then fin...

A---B

@Yashas Can I know the problem ?

Is JEE some kind of olympiad ?

Yashas

8:16 AM

Mar 29 at 12:29, by Madhuchhanda Mandal

@SirCumference "The engineering test" which is regarded as the toughest entrance examination in the world

@A---B The problem was if U was a linear function of T for all substances

Anonymous

@Yashas It is.

Yashas

nope

fusion of a substance

Anonymous

thermopedia.com/content/881

Anonymous

@Yashas Are you crazy? Internal energy increases with temperature man!

Yashas

nope

Feynman can't be wrong

Anonymous

8:24 AM

Let us ask John Rennie

Yashas

wait

U increases with T

but U can also increase without increase in T

Anonymous

@Yashas That's a different point

Yashas

:|

Anonymous

I am saying U is function of both T and V

Yashas

it tells that U is not a linear func of T

ok I messed up the words

hmm

Anonymous

8:26 AM

@Yashas Not "only" a function of T

Yashas

yea

I messed up the wording

Anonymous

Right

Anonymous

8:46 AM

@Yashas Is the change in U positive negative or zero when water freezes ? How will you find its value?

Anonymous

T remains constant

Yashas

dU = q + W

heat will leave the system

you won't do any work

internal energy decreases

Anonymous

Exactly. So U is a function of volume also!

Anonymous

:-)

Yashas

u do a little work

Anonymous

8:48 AM

W is not 0

Yashas

but it is near zero

Anonymous

That's right

Anonymous

So we have proved that U is function of V and T both :-)

Anonymous

$dU=nC_vdT + \frac{dU}{dV}dV$

Anonymous

$dT=0$

Anonymous

8:50 AM

$dU/dV$ is the internal pressure

Anonymous

$dV$ is positive

Anonymous

So internal pressure must decrease when water freezes

Anonymous

No wait

Anonymous

That doesn't make sense

Anonymous

@Yashas Why is $$(\frac{\partial U}{\partial V})_T$$ negative?

Yashas

8:52 AM

if volume has to increase, u need to do work

the work comes from the internal energy

as V increases, U decreases

Anonymous

@Yashas work can come from heat also..

Anonymous

dq=du+dw

Yashas

volume increases, pressure decreases

pressure = internal energy per unit volume

U = 2/3P

for that works for ideal gases only

Anonymous

$$dU=nC_vdT + \frac{\partial U}{\partial V}dV$$

Anonymous

This formula is true for all processes

Anonymous

8:55 AM

Not only for ideal gas

Anonymous

Change in Internal energy per unit change in volume must be negative at constant temperature during water freezing But why?

Anonymous

9:08 AM

Hmm. I think I got it.

Anonymous

$$P_{in}=\left|\left(\frac{\partial U}{\partial V}\right)_T\right|$$

Anonymous

9:20 AM

@Yashas If equal volumes of two ideal gases are mixed at constant temperature and pressure in an isolated container. Then what will be the signs of dU, dq and dW? This is confusing !!

Anonymous

@SmitChaudhary Hi

Anonymous

@Yashas I think dq=0

Anonymous

dq=dU+dW

Anonymous

For ideal gases U is only function of temperature

Anonymous

So dU=0

Anonymous

9:23 AM

dq=dU+dw....and hence dw=0....

Anonymous

I'm not sure about dw though

Yashas

9:39 AM

ok back

I was answering a Q on gamedev

Anonymous

10:11 AM

0

Q: Mixing of equal volumes of two ideal gases at constant temperature and pressure in an isolated container

"Mixing of equal volumes of two ideal gases at constant temperature and pressure in an isolated container" In this process why is net work done by the gases $0$ ? If we think about the left hand side container then it expands into a volume of $2V$ from $V$. So the work done must be positive fo...

thermodynamics

Anonymous

@Yashas answer that

Yashas

:|

answering

ETA 5 mins

Anonymous

You can answer it here also

Anonymous

:-P

Anonymous

@Yashas What?

Yashas

10:13 AM

estimated time

n is same for both

actually ur Q is unclear

Anonymous

@Yashas Yes, number of moles is same for both

Yashas

are the gases same?

Anonymous

Check out the jee 2015 paper

Yashas

I remember that Q

Anonymous

@Yashas No

Yashas

10:15 AM

it appears in chemistry, right?

Anonymous

Yes

Anonymous

@Yashas Got it?

Anonymous

@SwapnilDas Hello

Swapnil Das

Hi! :)

Yashas

Anonymous

10:23 AM

@Yashas But where is the explanation as to why work is 0? That toh even I could do. dq=0 and dU=0 so dw=0 from FLT

Yashas

Work is done during the transition

Anonymous

I wanted the logical explanation as to why net work done by gases in both compartments is 0

Yashas

but they are exchanging work with each other

ok let me add that

Anonymous

@Yashas Wait, what?

Anonymous

Exchanging ?

Anonymous

10:25 AM

Ah, partial pressure is continuously changing for each gas!

Anonymous

But that isn't good enough :/

Yashas

-_-=

Anonymous

I'm saying stupid things

Yashas

that Q will be, imo, closed

Anonymous

@Yashas Did you find the answer yet?

Anonymous

10:31 AM

I don't understand what you mean by exchanging work...

Yashas

During the process of transition, the gas on one side is doing work on the other. When the equilibrium is established, the work they did on each other does not matter. It does not affect the internal energy of the gas in the system as a whole. They were only exchanging energy between themselves.

I am adding some microscopic explanation

Anonymous

@Yashas That is again diverting from the point. I want to know why net dw is 0.

Anonymous

"the work they did on each other does not matter"

Yashas

:|

Anonymous

This is just evading the point ^

Yashas

10:32 AM

A does work on B

Anonymous

:(

Yashas

B's internal energy increases

B does work on A

When B's internal energy increases, B does work on A

Anonymous

A does negative work on B

Yashas

when A's internal increases, A does work on B

Anonymous

And B does negative work on A

Yashas

10:33 AM

wtf

A's internal energy increases

I am not sure what you are asking

can u be more clear?

Anonymous

I'm using the physics convention

Yashas

A does work on B and B does negative work on A is not meaningful.

Well, it is meaningful but you cannot add them up.

Either use work or use internal energy, not both.

Anonymous

Forget internal energy

Anonymous

I'm trying to explain using work done

Yashas

ok explain again

Anonymous

10:35 AM

So A is expanding and gas in B is doing negative work against A

Anonymous

Ok so far?

Yashas

ok

Anonymous

And similarly B is expanding and gas in A is doing negative work against B

Yashas

yes

Anonymous

So say $W_1$ and $W_2$ are the two works

Anonymous

10:36 AM

Then W1+W2<0

Anonymous

But that is not true

Anonymous

Why?

Anonymous

W1+W2=0 using FLT

Anonymous

The two explanations are giving different results

Yashas

what

you are doing summation of zeros

and the no of steps is infinite

W1 is greater than W2

but W3 done by A is going to be smaller than W4 done by B

they keep alternating

if u give infinite time, their sum will be equal

Anonymous

10:39 AM

@Yashas Yes, their sum will be equal with same sign

Anonymous

The total work won't be 0

Yashas

What?!

Anonymous

Say work done by A on B is -5 then work done by B on A will also be -5

Anonymous

Then total work will be -5-5=-10

Yashas

you are adding those two?

Anonymous

10:40 AM

And not zero

Yashas

You can't do that

they are different systems

Anonymous

@Yashas Why can't I added them ? The two combine to form a single system...

Yashas

^ you took them as two separate systems

you can't do both!

you can't assume them to be two separate and the same thing at the same time

This is funny and stupid.

Anonymous

Hmm, when you are finding $dq$ and $du$ aren't you assuming them to be a single system?

Yashas

part of the system did work W1 on another part of the system

the other part of the system did work W2 on the first part of the system

free energy

the system did net work

@blue No

I take the initial energy of both

then equate it with the final energy

@blue You have two like charges A and B; A is doing positive work on B and B is doing positive work on A; their sum is not zero!

you simply can't add like that

Anonymous

10:44 AM

@Yashas Yes, when you bring them together their potential energy increases

Yashas

yes but W1 + W2 = twice the decrease in potential energy

Anonymous

Right, so internal energy is in play here.

Yashas

that doesn't matter; your work argument is just wrong

Anonymous

@Yashas The internal energy of gas A remains same after the event, does it?

Yashas

yes

its volume increased and its pressure decreased

U = 2/3P where U is energy per unit volume

:O

Anonymous

10:52 AM

But in case of the charged spheres, when the two repel, if A does positive work on B then the potential energy of B decreases. As a whole dW-dU=0 So external work done is said to be 0.

Anonymous

I can't extend that logic to this case :/

Anonymous

I'm missing something

Yashas

When A does positive work on B, the internal energy of B increases. As a whole dW-dU=0

Anonymous

@Yashas Just now you said that internal energy remains same for gas A and B

Anonymous

look up ^

Yashas

10:55 AM

event?

I took that as the whole process

for the tiny event where A does work B, the internal energy changes

Anonymous

@Yashas It changes? Right? Then finally how come the internal energy of gas A remain same as before?

Anonymous

I'm confused now

Yashas

pff

let W_1a be the work done by gas A on gas B in the first step

let W_1b be the work done by the gas B on gas A in its first step

W_1a causes an increase in U_b

W_1b causes an increase in U_a

when U_b increases, the P_a increases

so becaz W_1a happened, W_1b happens

A does work on B; B does work on A; A does work on B; B does work on A; ....

A does work on B; B got compressed, it's pressure increased, therefore, B does work on A; Now A's P increased slightly, so A does work on B; ....

this keeps happening until they fully diffuse into each other

this also gives rise to two problems:

1. Which one starts? We don't care; one side starts.

2. Do they ever mix if they keep doing work on each other?

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